Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{-z^2 - 6z}{-3z - 15} \div \dfrac{3z^3 + 33z^2 + 30z}{3z^2 + 45z + 150} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-z^2 - 6z}{-3z - 15} \times \dfrac{3z^2 + 45z + 150}{3z^3 + 33z^2 + 30z} $ First factor out any common factors. $n = \dfrac{-z(z + 6)}{-3(z + 5)} \times \dfrac{3(z^2 + 15z + 50)}{3z(z^2 + 11z + 10)} $ Then factor the quadratic expressions. $n = \dfrac {-z(z + 6)} {-3(z + 5)} \times \dfrac {3(z + 10)(z + 5)} {3z(z + 10)(z + 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-z(z + 6) \times 3(z + 10)(z + 5) } {-3(z + 5) \times 3z(z + 10)(z + 1) } $ $n = \dfrac {-3z(z + 10)(z + 5)(z + 6)} {-9z(z + 10)(z + 1)(z + 5)} $ Notice that $(z + 10)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-3z\cancel{(z + 10)}(z + 5)(z + 6)} {-9z\cancel{(z + 10)}(z + 1)(z + 5)} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $n = \dfrac {-3z\cancel{(z + 10)}\cancel{(z + 5)}(z + 6)} {-9z\cancel{(z + 10)}(z + 1)\cancel{(z + 5)}} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $n = \dfrac {-3z(z + 6)} {-9z(z + 1)} $ $ n = \dfrac{z + 6}{3(z + 1)}; z \neq -10; z \neq -5 $